Last week was about deferred potentials. Basically you need to calculate the deferred potential in order to deal with time-dependent charge distributions.

The goal of today is to show you two not-very-rigorous derivativations of the intensity from a radiating dipole. Griffiths does the rigorous derivativation, but I want to show you the forest for the trees. After we talk about the forest today, hopefully you’ll have more appreciation for the trees when you read section 11.1.

Here’s what the deferred potentials look like

\[V(\vec{r}, t) = \frac{1}{4\pi\epsilon_0}\int\frac{\rho(\vec{r}^\prime, t_d)}{\mathscr{r}}d\tau^\prime\] \[\vec{A}(\vec{r}, t) = \frac{\mu_0}{4\pi}\int\frac{\vec{J}(\vec{r}^\prime, t_d)}{\mathscr{r}}d\tau^\prime\]

The deferred time corresponds to:

\[t_d = t - \frac{\mathscr{r}}{c}\]

Here are the Leinard-Wiechert potentials for a moving point charge.

\[V(\vec{r},t) = \frac{1}{4\pi \epsilon_0}\frac{qc}{(\mathscr{r}c - \vec{\mathscr{r}}\cdot\vec{v})}\]

where \(\vec{v}\) is the velocity of the charge at the deferred time, and \(\vec{r}\) is the vector from the deferred position to the place you’re calculating the field \(\vec{r}\).

The vector potential is nice and simple:

\[\vec{A}(\vec{r},t) = \frac{\vec{v}}{c^2}V(\vec{r},t)\]

Last class I spent awhile motivating that difference, i.e. motivating why the different between c and v comes up in the equations.

We considered a moving train and I convinced you that the length of the train in the moving frame was:

\[L^\prime = \frac{L}{1 - \frac{v}{c}}\]

So the train appears longer by a factor of \(\frac{1}{1 - \frac{v}{c}}\) Once you allow for the train not to be coming straight toward you, that v/c turns into a \(\hat{\mathscr{r}}\cdot\vec{v}/c\)

This translates into a decreased charge density (because of increased volume) for charges moving relative to you. And that’s where that difference in the denominator of V comes from.

Radiation

Punchline and main point: when charges accelerate their fields transport energy irreversibly to infinity. That transport of energy to infinity is called radiation.

We have studied power in a field. The Poynting vector is power per area. So power is the Poynting vector \(S\) integrated over an area.

\[P = \int \vec{S}\cdot d\vec{a} = \frac{1}{\mu_0}\int(\vec{E}\times\vec{B})\cdot d\vec{a}\]

If this radiation is going to be transported to infinity then we need it to actually leave. Let’s think about it leaving some area that we designate. How much power exits some volume bounded by a closed surface? (I just made the previous integrals closed surface integrals.)

\[P = \int \vec{S}\cdot d\vec{a} = \frac{1}{\mu_0}\int(\vec{E}\times\vec{B})\cdot d\vec{a}\]

Well if we want this power to actually leave the sphere, and also to go to infinity, then no matter what size sphere we choose, the power leaving it should be the same. (Imagine a light bulb emitting 60 watts of power. There should be 60 watts of power measured at any distance as longas you integrate over the sphere. Otherwise energy is bunching up someplace.)

So then the part of this pointing vector that correponds to radiation is the part where \(E\times B\) decreases at \(1/r^2\). Otherwise we’ll measure different amounts of power at different radii.

That means that we only want the part of the field that drops off as \(1/r\). That’s not the coulomb field. The coulomb field of a point charge drops off as \(1/r^2\). So that’s not transporting energy.

Let’s look at two equations we didn’t talk about last time. They are called Jefimenko’s equations and they result directly from the deferred potential above. Let’s just write down E.

\[E(\vec{r}, t) = \frac{1}{4\pi\epsilon_0}\int\left[\frac{\rho(\vec{r}^\prime, t_d)}{\mathscr{r}^2}\hat{\mathscr{r}} +\frac{\dot\rho(\vec{r}^\prime, t_d)}{c\mathscr{r}}\hat{\mathscr{r}} - \frac{\dot{\vec{J}}(\vec{r}^\prime, t_d)}{c^2\mathscr{r}}\right]d\tau^\prime\]

That first term is the Coulumb term (and the whole thing reduces to the columb field if nothing has any time dependence.) And the second two terms are the ones that decrease with r. The first and second terms came from using the chain rule in differentiation \(\rho/\mathscr{r}\). The third term came from the new thing we added to the field - we don’t just need \(\nabla V\) anymore. Now we also need \(\frac{\partial \vec{A}}{\partial t}\).

Dipole radiation

So it turns out that calculating the radiation from an oscillating dipole is pretty easy. I want you to guess what the formula is. Just using dimensional analysis (not using Jefimenko’s formulae).

You know the potential of a dipole: (We’re really starting to pull a lot of things together now.)

\[V = \frac{1}{4\pi\epsilon_0)} \frac{p \cos\theta}{r^2}\]

That’s for a dipole in the z direction, with \(\theta\) being the polar coordinate.

Come up with a formula for intensity (power per area), that includes \(p, \omega, \epsilon_0, \mu_0, c\) and \(r\) where \(\omega\) is the frequency at which the dipole is oscillating. You know these things:

  • The poential of a dipole (above).
  • E is the spatial derivative of V.
  • B is E/c
  • Intensity is S given by \(\frac{1}{\mu_0}E\times B\).
  • Intensity must decrease by \(1/r^2\).

Ans: I got that the units of \(<S>\) which I will designate \([S]\) must be

\[[S] = \left[\frac{1}{\mu_0}\right][E][B] = \left[\frac{1}{\mu_0\epsilon_0^2 c}\frac{p^2}{r^6}\right]\]

And then we were told it has to go like \(1/r^2\) so I need to convert \(r^4\) into c’s and \(\omega's\). So then I get

\[<S> \propto \frac{\mu_0}{c}\frac{p_0^2\omega^4}{r^2}\]

That actually is the formula down to a factor of \(32\pi^2\) which admittedly is pretty big. But that’s all the dependencies. I wrote the dipole moment \(p\) as \(p_0\) to indicate that \(p_0\) is the amplitude of the oscillating dipole.

More of a derivation, but still pretty cavalier

Imagine that we have two small spheres connected by a wire, and q(t) is the charge on the top sphere. Here’s how we write down the charge.

\[q(t) = q_0 \cos{\omega t}\]

We know that the radiation depends on dA/dt (among other things) from staring at Jefimenko’s equation, so let’s rely on A to estimate this.

A involves J, so let’s calculate I.

\[I(t) = \frac{dq}{dt}= -q_0 \omega \sin{\omega t}\]

To get A we normally integrate J over a volume, but I is all happening in one dimension, so integrating J over a volume is the same thing as integrating I over a length. (Remember J is I per area.)

The integral would look something like this: \(\vec{A} = \frac{\mu_0}{4\pi} \int\frac{\vec{J}(\vec{\mathscr{r}})}{\mathscr{r}}d\tau^\prime\)

\[\vec{A} = \frac{\mu_0}{4\pi} \int\frac{\vec{I}}{r}dz^\prime\]

We’ll say our dipole is small enough that \(\mathscr{r}\) is basically just \(r\). Therefore the integral doesn’t depend upon it. Formally we’re saying the linear extent of the dipole is small compared to the distance we are away. So then..

\[\vec{A} = \frac{\mu_0}{4\pi} \frac{I d}{r}\]

where I integrated dz over the length of the dipole, d (the separation of the charges).

So now \(A = -\frac{\mu_0}{4\pi} \frac{q_0d\omega \sin{\omega t}}{r}\)

\[\frac{\partial {A}}{\partial t} = -\frac{\mu_0}{4\pi} \frac{q_0d\omega^2 \sin{\omega t}}{r}\]

Oh yay! We’ve still got 1/r! That’s what we wanted. And now we’ve got everything we need to calculate \(S\).

\[S = |\frac{1}{\mu_0}(\vec{E}\times \vec{B})| = \frac{1}{\mu_0 c} E^2\] \[S = \frac{\mu_0 p_0^2 \omega^4}{16\pi^2 c r^2} \sin^2\omega t\]

where I let \(p_0 = qd\).

The only other piece we need is that time-averaged amplitude of a sine-wave squared is 1/2 the amplitude of it. So the formula for the time-averaged intensity of dipole radiation is

\[<\vec{S}> = \frac{\mu_0 p_0^2 \omega^4}{32\pi^2 c} \frac{\sin^2\theta}{r^2} \hat{r}\]

The dependence on \(\theta\) is really important. You don’t get any radiation looking down on the dipole.

The \(\sin \theta\) comes from decomposing \(\hat{z}\) the direction of the dipole, into \(\hat{theta}\) and \(\hat{r}\). The \(\cos\theta \hat{r}\) term goes away because it ends up cancelling with the \(\theta\) term in the scalar potential.